Problem: Simplify and expand the following expression: $ \dfrac{2}{2r - 16}+ \dfrac{2}{2r + 2}+ \dfrac{4r}{r^2 - 7r - 8} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{2}{2r - 16} = \dfrac{2}{2(r - 8)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{2}{2r + 2} = \dfrac{2}{2(r + 1)}$ We can factor the quadratic in the third term: $ \dfrac{4r}{r^2 - 7r - 8} = \dfrac{4r}{(r - 8)(r + 1)}$ Now we have: $ \dfrac{2}{2(r - 8)}+ \dfrac{2}{2(r + 1)}+ \dfrac{4r}{(r - 8)(r + 1)} $ The least common multiple of the denominators is: $ 4(r - 8)(r + 1)$ In order to get the first term over $4(r - 8)(r + 1)$ , multiply by $\dfrac{2(r + 1)}{2(r + 1)}$ $ \dfrac{2}{2(r - 8)} \times \dfrac{2(r + 1)}{2(r + 1)} = \dfrac{4(r + 1)}{4(r - 8)(r + 1)} $ In order to get the second term over $4(r - 8)(r + 1)$ , multiply by $\dfrac{2(r - 8)}{2(r - 8)}$ $ \dfrac{2}{2(r + 1)} \times \dfrac{2(r - 8)}{2(r - 8)} = \dfrac{4(r - 8)}{4(r - 8)(r + 1)} $ In order to get the third term over $4(r - 8)(r + 1)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{4r}{(r - 8)(r + 1)} \times \dfrac{4}{4} = \dfrac{16r}{4(r - 8)(r + 1)} $ Now we have: $ \dfrac{4(r + 1)}{4(r - 8)(r + 1)} + \dfrac{4(r - 8)}{4(r - 8)(r + 1)} + \dfrac{16r}{4(r - 8)(r + 1)} $ $ = \dfrac{ 4(r + 1) + 4(r - 8) + 16r} {4(r - 8)(r + 1)} $ Expand: $ = \dfrac{4r + 4 + 4r - 32 + 16r}{4r^2 - 28r - 32} $ $ = \dfrac{24r - 28}{4r^2 - 28r - 32}$ Simplify: $ = \dfrac{6r - 7}{r^2 - 7r - 8}$